Question

Solve $6{{x}^{3}}-11{{x}^{2}}+6x-1=0$ , if its roots are in harmonic progression.

Answer 1

Hint: Think of the basic definition of Harmonic progression, and using the given equation, try to reach an equation whose roots are in arithmetic progression.

Complete step by step answer:
Before starting with the solution to the above question, let us talk about harmonic progression.
In mathematics, harmonic progression is defined as a sequence of numbers, for which the sequence of the reciprocals of its terms gives an arithmetic progression.
We can represent a general harmonic progression as:$\dfrac{1}{a},\dfrac{1}{a+d},\dfrac{1}{a+2d}............$
Where $a,a+d,a+2d...............$ will represent an arithmetic progression.
Now, let us start with the solution to the above question.
The equation given in the question is an example of a cubic polynomial whose roots are in harmonic progression. Therefore, substituting x in the given cubic polynomial with $\dfrac{1}{t}$ will give a polynomial whose roots are in arithmetic progression. So, substituting x by $\dfrac{1}{t}$ , we get
$6{{x}^{3}}-11{{x}^{2}}+6x-1=0$
$\Rightarrow \dfrac{6}{{{t}^{3}}}-\dfrac{11}{{{t}^{2}}}+\dfrac{6}{t}-1=0$
Now we will multiply both sides of the equation by ${{t}^{3}}$ , on doing so our equation becomes:
${{t}^{3}}\times \dfrac{6}{{{t}^{3}}}-{{t}^{3}}\times \dfrac{11}{{{t}^{2}}}+{{t}^{3}}\times \dfrac{6}{t}-{{t}^{3}}\times 1={{t}^{3}}\times 0$
$\Rightarrow -{{t}^{3}}+6{{t}^{2}}-11t+6=0$
Now, as we mentioned, the roots of the equation $-{{t}^{3}}+6{{t}^{2}}-11t+6=0$ is in arithmetic progression, we let the root to be ( a - d ), a, ( a + d ).
Therefore, using the relation between the coefficients and roots of a polynomial, we get
 Sum of roots of a polynomial = $\dfrac{\text{-}\left( \text{coefficient of }{{\text{x}}^{\text{2}}} \right)}{\left( \text{coefficient of }{{\text{x}}^{\text{3}}} \right)}$
$\therefore a-d+a+a+d=\dfrac{-6}{-1}$
$\Rightarrow 3a=6$
$\Rightarrow a=2...........(i)$
Now, product of roots of a cubic polynomial taken all at a time = $\dfrac{\text{-}\left( \text{constant} \right)}{\left( \text{coefficient of }{{\text{x}}^{\text{3}}} \right)}$
$\therefore a\left( a-d \right)(a+d)=\dfrac{-6}{-1}$
Now we will substitute the value of ‘a’ from equation (i), we get
$2\left( 2-d \right)(2+d)=6$
We know $\left( x+y \right)(x-y)={{x}^{2}}-{{y}^{2}}$ . So, our equation becomes:
 $2\left( {{2}^{2}}-{{d}^{2}} \right)=6$
$\Rightarrow 4-{{d}^{2}}=\dfrac{6}{2}$
$\Rightarrow 4-{{d}^{2}}=3$
$\Rightarrow 4-3={{d}^{2}}$
$\Rightarrow {{d}^{2}}=1$
Taking square root of both sides of the equation, we get
$\sqrt{{{d}^{2}}}=\sqrt{1}$
$\therefore d=1$
Therefore, the roots of the equation $-{{t}^{3}}+6{{t}^{2}}-11t+6=0$ are ( a - d ), a, and ( a + d ) which comes out to be 1, 2, and 3.
We also know that the roots of the equation $6{{x}^{3}}-11{{x}^{2}}+6x-1=0$ is reciprocal of the roots of the equation $-{{t}^{3}}+6{{t}^{2}}-11t+6=0$ .
Therefore, the roots of the equation $6{{x}^{3}}-11{{x}^{2}}+6x-1=0$ is 1, $\dfrac{1}{2}$ and $\dfrac{1}{3}$ .

Note: If you want, you can directly solve the given equation without substituting x by $\dfrac{1}{t}$ , but that might make the equations complicated to solve. As to find the roots of the equation $6{{x}^{3}}-11{{x}^{2}}+6x-1=0$ directly you need to let the roots be $\dfrac{1}{a},\dfrac{1}{a+d}\text{ and }\dfrac{1}{a+2d}$ which would eventually lead you to a cubic equation when you apply the relation between the coefficients and roots of a polynomial.